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The Natural
By
Ian Beardsley
Copyright © 2022 By Ian Beardsley!
of 2 58
Contents
1.0 Atoms As A Vibrating String……………………………………….4
2.0 Time as Rhythm………………………………………………………..6
3.0 Reconciling Gravitational Fields With Electric Fields……9
4.0 How the Elements Organize……………………………………….14
Appendix 1……………………………………………………………………..12
5.0 Giordano’s Relationship And Proton-Seconds………25
Appendix 2……………………………………………………………..52
Appendix 3……………………………………………………………..53
Appendix 4……………………………………………………………..55
6.0 Bringing It All Together……………………………………..57
of 3 58
The distribution of two distinct tones on the Middle Eastern hand-drum over 4 beats results in
several permutations for which so-called Middle Eastern four is the basis set that defines a
meter of four for which all other sets in the same meter, such as Baladi and Chiftatelli, are
variations. This set is Natural and describes Nature by way of its syncopation (playing of two
over three and three over four). Similarly, insofar as music, is the playing of notes (frequencies)
over time described by these rhythms, we see the physical aspect of atoms can be described as
vibrating frequencies for which the mass, tension, and length of the string determines its states
in that a vibrating string has frequency (cycles per second) and an atom has frequency as cycles
per second of the orbit of an electron around a proton and its associated emission and
absorption spectra which are colors of light defined by their frequencies as electromagnetic
radiation (light). In terms of this an approach to a single equation for gravitational fields and
electric fields is approached. It is suggested that there is perhaps no relationship between the
physical causes of electric and gravitational fields but perhaps there is between their respective
behaviors for the organization of the elements in the periodic table such that life can arise,
which leads to a treatment of the elements as mathematical patterns. Curious constants both
equal to six in units we thought were arbitrary: Giordano’s Relationship is (6kg2)(s/m). Proton-
seconds is six proton-seconds. They seem to be connected to one another, and relate the
microcosmos to the macrocosmos (atoms to solar system). We predict the hydrocarbons which
are the skeleton of life chemistry, the radius of a proton, the radius of the solar system and the
abundances of the primordial elements in the Universe.
of 4 58
1.o Atoms As A Vibrating String To describe the atom as a vibrating string as on any
musical instrument, we begin with the force between the charges of the proton and the electron
in a hydrogen atom which is given by Coulomb’s law:
Equation 1.
The separation between the proton and electron is at any time most probabilistic to be the Bohr
radius, which is given by:
Equation 2.
Where is the permittivity of free space, is the reduced Planck constant, and is
the elementary charge. is the electron rest mass. The reduced Planck Constant is the Planck
constant divided by . Letting we have and letting F be the the tension, T, in the
string, we have:
Equation 3.
Where we have let Coulomb’s constant k to equal . The fine structure constant squared is
the ratio of the potential energy of an electron in the first circular orbit to the energy given by
times the speed of light squared. That is:
Equation 4.
We will say this potential energy is equal to the kinetic energy of the electron which is in
turn given by the frequency of the the vibration of our string, which is the frequency of the
electron. The energy of the electron in terms of frequency f, is . We have:
Equation 5.
Equation 6.
The velocity of a vibrating string is given by the tension of the string, T, and the mass of the
string per unit length as the square root of their ratio:
Equation 7.
Combining equation 4, 5, 6, and 7 we have:
F = k
q
1
q
2
r
2
a
0
=
4πϵ
2
0
ℏ
2
m
e
e
2
ϵ
0
ℏ
e = q
1
= q
2
m
e
h
2π
r = a
0
T =
m
2
e
e
6
π
4 ϵ
3
0
h
4
1
4πϵ
0
U
e
m
e
α
2
=
U
e
m
e
c
2
1
2
m
e
v
2
E = h f
h f =
1
2
mv
2
h f = α
2
m
e
c
2
μ
v =
T
μ
of 5 58
Equation 8.
This would be the mass per unit length of the vibrating string that represents hydrogen with the
electron in its first orbital. We see this value is:
Equation 9.
The radius of a hydrogen atom is 1.2E-10m. Thus this string has a mass of 2.868E-39kg. The
mass of a proton is 1.67262E-27kg. This makes it 5.8320E11 times more massive. The mass of an
electron is 9E-31 making it 3.138E8 times more massive. The ratio of the mass of a proton to
that of an electron is 1.858467E3 making it 1.858467E3 times more massive (1,858.5 times more
massive).
μ =
π
8α
2
⋅
m
e
e
6
c
2
ϵ
3
0
h
4
μ = 2.39 × 10
−29
kg/m
Both sound and atoms can
be characterized by waves.
of 6 58
2.0 Time as Rhythm
of 7 58
of 8 58
We can write our matrix as follows…
maps where
and
This means the first equation is 2 full and the second equation is 1 full, which is correct in terms
of our drawing of doums and teks distributing themselves around a circle over time. We could
make our matrices differently and describe these distributions over time as 2 Full 2 Empty, 1
Empty 1 Full, or 1 Empty 2 Empty…
It doesn’t matter which one you choose, as along as you choose one and address it accordingly. It
is that old adage: “Is the glass half-empty or is it half full?”
T = A
x
ℝ
3
→ ℝ
2
A =
(
−sinθ
1
sinθ
2
−sinθ
1
cosθ
2
−sinθ
1
−cosθ
1
)
A
x =
b
A
x =
(
−sinθ
1
sinθ
2
−sinθ
1
cosθ
2
−sinθ
1
−cosθ
1
)
x
1
x
2
x
3
=
(
b
1
b
2
)
x
1
= x
2
= x
x
= 1
θ
1
= −
π
2
θ
2
= − π
−x
1
sinθ
1
+ x
2
sinθ
2
− x
3
sinθ
1
= b
1
x
1
cosθ
2
− x
2
sinθ
1
− x
3
sinθ
1
= b
1
(1) − (0) + (1) = 2
() − (−1) + (0) = 1
A = − si nθ
1
i + sinθ
2
j − sinθ
1
k
B = cosθ
2
i − sinθ
1
j − cosθ
1
k
of 9 58
=(1, 0, 0)+(0, 0, 1)=(1, 0, 1)
And we see with our matrix as we have organized it is at an angle 45 degrees.
3.0 Reconciling Gravitational Fields With Electric Fields
One of the comparisons I come across which doesn’t make sense to me is that the electric forces
are many times stronger the gravitational forces. The comparison is done by computing the
gravitational force of a proton by inputting its mass in:
And comparing this too the electric force produced by the mass of a proton with
But you will notice while both are inverse square fields one is a function of charge q, and the
other of mass m. There is nothing here that says charge is a function of mass and mass is a
function of charge. It is comparing apples and oranges. However at the same time we know that
these two equations cannot be reconciled with one another and that is why so many theoretical
physicist are looking for a grand unified field theory, one equation that produces both forces. I
think to do this we need a common cause between the production of these different forces. I
have described the the atom with its electron coupled to its proton by a vibrating string with
mass per unit length and tension T. That which we should seek here is the gravitational force as
coupling two masses with a vibrating string as well.
We start with the Universal constant of gravitation:
N=newtons=force is the tension of our string, T. We see that
i
j
k
−sinθ
1
sinθ
2
−sinθ
1
cosθ
2
−sinθ
1
−cosθ
1
= −
i +
k
F = G
m
1
m
2
r
2
F = k
q
1
q
2
r
2
G = 6.67408 × 10
−11
N ⋅
m
2
kg
2
of 10 58
Take the square root of both sides:
The velocity of a string with tension T and mass is
But, we have
We suppose there is some x for which this holds, that is such that:
What I would like to say here is that the possibility exists that electric and gravitational forces
did not spring in their origins from a common force, that there is no relationship between force
as a function of charge and force as a function of mass because they are distinct and
independent ideas as far as physical cause goes. But this is not necessarily a bad thing because it
could be that charge is related to mass in terms of something other than being unified fields and
is rather the degree of charge produces the forces in atoms such that the elements form in a way
that under gravity in the formation of the protoplanetary disc, that the elements have properties
that give them their relative abundances in the composition of the planets such that life could
arise (for example in that the earth would have the three quarters of its surface covered in water
in its liquid phase that it has). This idea is not less in aesthetic elegance than the former but is
simply a characteristic that might be in play and exists abstractly in mathematics; there are two
types of systems in linear algebra, linearly dependent and linearly independent. Consider the
matrix
It is in echelon form:
G ⋅
kg
2
m
2
= T
G
kg
m
= T
μ = kg/m
v =
T
μ
G =
T
μ
Gx =
T
μ
x =
T
μ
1
G
=
kg
m
1 1 h
1 h 1
h 1 1
c
1
c
2
c
3
=
(
0
0
0
)
of 11 58
The system is linearly independent if h is anything but -2 and +1. The search for a unified field
theory is like looking for an h in nature such that the systems of equations can be linearly
dependent. That h may or may not exist. This leads us however to another endeavor, if the earth
has the elements in their relative abundances that it has, and that has function, then the
elements are organized in the periodic table such that they are organized as mathematical
patterns. That is the subject of the next section.
1 1 h
0 h − 1 1 − h
0 0 2 − h − h
2
c
1
c
2
c
3
=
(
0
0
0
)
of 12 58
Appendix 1
of 13 58
of 14 58
4.0 How the Elements Organize
Silicon (Si) has four valence electrons and wants 8 to complete an octet. We can dope it with
boron (B) which donates three, leaving a deficit, or hole which can carry current for positive or
p-type silicon. Or, we could dope it with phosphorus which donates 5 giving it a surplus that can
carry a current making negative or n-type silicon. The two types can be joined to make a
semiconductor, or gate through which current only travels in one direction. As such we can
make diodes and transistors to create a binary network of decision making nodes, a computer.
We can do the same with germanium (Ge) and doping agents Ga and As, which can use with
silicon as well. We can call these the AI elements and we notice they can be pulled directly out of
the periodic table of the elements to make a 3 by 3 matrix, and we can take the cross product or
even the curl. We have
Which is:
=
=
We can dot it with the following area elements
We have
=
i
j
k
∂
∂x
∂
∂ y
∂
∂z
( − C ⋅ P)y (Si ⋅ G a)z (G e ⋅ As)y
(Ge ⋅ As − Si ⋅ G a)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
d
a =
(
zd yd z
i + yd yd z
k
)
∇ ×
u = (Ge ⋅ As − Si ⋅ G a)
i + (C ⋅ P)
k
(Ge ⋅ As − Si ⋅ G a)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
of 15 58
=
=154,082,837.980+16,452,521.6822=
The end result is
Equation 10:
We can also form a 3 by 3 matrix with the electronics elements of the finest electrical, ductile,
malleable conductors copper (Cu) silver (Ag) and gold (Au).
We can remove the 5th root sign in equation 1 by noticing
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
)
⋅
(
zd yd z
i + yd yd z
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
⋅ zd zd y + 372.025855
(
g
m ol
)
2
⋅ yd zd y
)
∫
Ge
Si
3,484 . 134569
(
(72.64 − 28.085)
2
2
)
d y +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
m ol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
m ol
)
3
(
(72.64 − 28.085)
2
2
)
170,535, 359.662
(
g
m ol
)
5
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = ex p
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
of 16 58
Equation 11.
Which is
(28.085)(72.64)(12.085)(107.8682)(196.9657)
Yielding:
Equation 12.
With carbon, the core biological life element separating semiconductor materials Si and Ge from
finest conductors Ag and Au. This works because
And, since
(Perimeter)/(Diameter of regular hexagon) = 3.00
The same value as our 3.0716 if taken at two places after the decimal.
Important
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the
elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and
germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain
noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four
electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope
the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two
5
∏
i=1
x
i
= Si ⋅ Ge ⋅ C ⋅ A g ⋅ Au = 523,818, 646.5
g
5
m ol
5
π
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
5
∏
i=1
x
i
523,818, 646.5
170,535, 359.662
= 3.0716
π = 3.141 . . .
3.141 + 3.00
2
= 3.0705
of 17 58
types of silicon we have a semiconductor for making diodes and transistors from which we can
make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the
right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.
I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy
in the differential because boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with
subscripts that tell what element they are and their properties. I have done this:
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the
first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be
organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in
three dimensional space so the mechanics of the realm we experience are described by such a
matrix, for example the cross product. Hence this paper where I show AI and biological life are
mathematical constructs and described in terms of one another.
We see if we include the two biological elements in the matrix (E14) and (E15) which are carbon
and nitrogen respectively, there is every reason to proceed with this paper if the idea is to show
not only are the AI elements and biological elements mathematical constructs, they are
described in terms of one another. We see this because the first row is ( B, C, N) and these
happen to be the only elements that are not core AI elements in the matrix, except boron (B)
which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means we have proved our case because the
first row if we take the cross product between the second and third rows are, its respective unit
vectors for the components meaning they describe them!
The Computation
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
A = (Al, Si, P)
B = (G a, Ge, As)
A ×
B =
B
C
N
Al Si P
Ga Ge As
= (Si ⋅ As − P ⋅ Ge)
B + (P ⋅ G a − Al ⋅ As)
C + (Al ⋅ G e − Si ⋅ Ga)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/m ol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/m ol
of 18 58
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological
elements C and N being the unit vectors are multiplied by the AI elements, meaning they
describe them! But we have to ask; Why does the first row have boron in it which is not a core
biological element, but is a core AI element? The answer is that boron is the one AI element that
is out of place, that is, aluminum is in its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)
is in boron’s place. This results in an interesting equation.
Equation 13.
The differential across germanium crossed with silicon plus the differential across silicon
crossed with germanium normalized by the product between silicon and germanium is equal to
the boron divided by the average between the germanium and the silicon. The equation has
nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the
equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the
harmonic mean between the semiconductor elements (by molar mass):
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsin θ = (50)(126)sin8
∘
= 877.79
877.79 = 29.6g /m ol ≈ Si = 28.09g /m ol
Si(A s − G a) + G e(P − Al )
SiGe
=
2B
Ge + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
of 19 58
Equation 14.
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
We can make this into two integrals:
Equation 15.
Equation 16.
If in the equation (The accurate harmonic mean form):
Equation 17.
We make the approximation
Equation 18.
Then the Stokes form of the equation becomes
Si
B
(A s − G a) +
Ge
B
(P − Al ) =
2SiGe
Si + Ge
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d x d y ≈
1
Ge − Si
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Si
B
(As − Ga)d ydz ≈
1
3
1
(Ge − Si)
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Ge
B
(P − Al )d x d z ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
of 20 58
Equation 19.
Thus we see for this approximation there are two integrals as well:
Equation 20.
Equation 21.
For which the respective paths are
By making the approximation
In
We have
Equation 22.
is the differential across Si, is the differential across Ge
and is the vertical differential.
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
dyd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)d ydz =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )dyd z =
2
3
∫
Ge
Si
dz
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
2SiGe
Si + Ge
≈ Ge − Si
Si(As − Ga)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − G a
ΔS = Ge − Si
of 21 58
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor elements by molar mass which are used
to make circuits.
We say (Phi) is given by
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 23.
Thus since
And we have
Equation 24.
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b /a
ϕ
(ϕ)ΔGe + (Φ)Δ Si = B
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔG e =
ΔS
Si
B −
G e
Si
ΔSi
of 22 58
We see and are both and c is in the Si (silicon) field wave, but for E and B fields c is the
speed of light.
To find the Si wave our differentials are
It is amazing how accurately we can fit these differentials with an exponential equation for the
upward increase. The equation is
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
G e
Si
= μ ϵ
0
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
G e = 0
ΔC = N − B = 14.01 − 10.81 = 3.2
ΔSi = P − Al = 30.97 − 26.98 = 3.99
ΔGe = A s − G a = 74.92 − 69.72 = 5.2
ΔSn = Bi − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 23 58
This is the halfwave….
Equation 25.
Interestingly, the 0.4 is boron (B) over aluminum (Al) the very two elements that lead us to
looking for a wave equation because boron was the out of place element in the AI periodic table
that lead to us to using aluminum as its dummy representative in the Si differential and that
itself divided into the left hand terms to give us the harmonic mean between the central AI
elements semiconductor materials Si and Ge. The Ag and Cu are the central malleable, ductile,
and conductive metals used in making electrical wires to carry a current in AI circuitry.
Conclusion
The biological elements and AI elements are mathematical constructs based
around pi and phi. Because 2/3 and 1/3 are approximations to and in
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
y(x) = e
B
Al
x
+
A g
Cu
B
Al
=
10.81
26.98
= 0.400667
A g
Cu
=
107.87
63.55
= 1.6974 ≈ 1.7
ϕ
(1 − ϕ)
of 24 58
And the in this is an approximation to
(Perimeter)/(Diameter of Regular Hexagon) = 3.00
∫
1
0
∫
1
0
Si
B
(As − Ga)d ydz =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d x d z ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
π
3.141 + 3.00
2
= 3.0705
π
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
5
∏
i=1
x
i
of 25 58
5.0 Giordano’s Relationship And Proton-Seconds
of 26 58
Giordano’s Relationship We now formulate what I call Giordano’s Relationship:
Warren Giordano writes in his paper The Fine Structure Constant And The
Gravitational Constant: Keys To The Substance Of The Fabric Of Space, March 21,
2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational constant
numerically, but neglecting any units.
Let’s do that!
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js!
And it works, G is:!
G=6.67408E-11 N(m2/kg2)!
Since Avogadro’s number is I suggest we formulate this as:!
Equation 26 !
Where and H=1 gram/atom!
Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so
on for 6E23 atoms per gram. Thus,…!
!
Since grams and atom cancel we can work in grams even though our equations are in
kilograms. Let us not write H, since formally it is grams per mole of hydrogen but write !
!
We have:!
!
Or,…!
Equation 27 !
Where!
h
1 + α
α
10
23
N
A
= 6.02E 23atom s/mole
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
N
A
H = 6.02E 23
atoms
gram
⋅
1gram
atom
= 6.02E 23
ℍ = 1
gram
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
of 27 58
Equation 28 !
Let us say we were to consider Any Element say carbon . Then in general!
Equation 29 !
We have!
and !
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have 12-6=6 grams of protons in the 12 grams of protons and neutrons.
Thus!
!
And it follows that!
!
We see in general since the atomic number Z is the number of protons in an atom that in
general this holds for all elements because!
!
And,!
!
Therefore we always have:!
Equation 30 !
This works nicely because we formulated molar mass nicely; we said element one (hydrogen)
which is one proton and one electron has one gram for a mole of atoms. Historically this was
done because we chose carbon (element six) to have 12 grams per mole, and determined what
the mole was such that it would hold. The reason this works is that hydrogen is one proton and
has no neutrons, but carbon has twelve neutrons but since hydrogen doesn’t have any
neutrons, and the neutron has the same mass as the proton, and our theory makes use only of
protons (in this instance of its formulation)!
x = 1.00kg
2
s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gram s
6protons
N
A
=
6(6E 23protons)
6gram s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23protons
Z ⋅ grams
𝔼 =
Z ⋅ grams
Z ⋅ protons
N
A
⋅ 𝔼 = 6E 23
of 28 58
!
comes out to have x equal to 1.00 (nearly) even. It is at this moment that we point out, because
it is important, that in equation!
!
is not molar mass, and that is a variable determined by ; A mole of protons multiplied
by the number of protons in . The reason we point this out, though it may already be clear, is
we wish to find the physical theory behind it. That is we need to find the physical explanation
for equation:!
!
It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in
kilograms, seconds, and meters, it may mean these units of measurement have some kind of a
meaning. We can in fact write it:!
!
We know that!
!
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed
of light squared. To begin our search for the meaning of equation 4 we convert x, the factor of
1.00 to astronomical units, years, and solar masses, as these are connected to the
orbit of earth as it relates to the sun. We have:!
!
= !
!
!
x = 1.00kg
2
s
m
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
kg
2
1
⋅
s
m
⋅
(1.98847E30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E33m
3
⋅
1.98847E30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
of 29 58
!
We can now write!
Eq 31 !
This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital
velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum
mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We
multiply both sides by and we have earth velocity on the left and the units stay the same
on the right. But what we will do is return to the form in kg-m-s and leave it as an equation but
put in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:!
Eq. 32 !
This brings up an interesting question: while we have masses characteristic of the
microcosmos like protons, and masses characteristic of the macrocosmos, like the minimum
mass for a star to become a neutron star as opposed to a white dwarf after she novas (The
Chandrasekhar limit) which is 1.44 solar masses, we do not have a characteristic mass of the
intermediary world where we exist, a truck weighs several tons and tennis ball maybe around a
hundred grams. To find that mass let us take the geometric mean between the mass of a
proton and the mass of 1.44 solar masses. We could take the average, or the harmonic mean,
but the geometric mean is the squaring of the proportions, it is the side of a square with the
area equal to the area of the rectangle with these proportions as its sides. We have:!
!
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is
. We have the intermediary mass is:!
Eq. 33 !
All we really need to do now is divide 7 by 8 and we get an even number that is the six of our
six-fold symmetry.!
Eq. 34 !
The six of our six-fold symmetry. We have something very interesting here. We have!
G ≈ 40
AU
3
M
⊙
⋅ year
2
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
4π
2
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
of 30 58
This is:
Equation 35.
Where k is a constant, given
Equation 36.
We can take the velocity of earth as being 30,000 m/s by rounding it. We have
We see every other planet is quantized as
an integer such that we produce the area
of its triangle is 6 (Fig. 1).
We also see our constant k is:
Eq. 37.
We see that in equation 7
That, is important because it suggests
the idea of
This is different than molar mass it is a gram
per proton is hydrogen 6 grams per six
protons is carbon…an so on. It is as if a gram
per proton is a natural constant. We see
Giordano’s Relationship connects protons to
the planets. My formulation of proton-
seconds did the same thing. Let’s look at that.
k v
e
= 6
k =
1
800
s
m
30,000
800
= 37
1
2
37.5 = 6.123734357
k =
1
m
2
i
⋅ h
(1 + α)
G
⋅ N
A
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
N
A
𝔼
value =
gra m
proton
of 31 58
Proton-Seconds It is as if the ancient cultures, tribes, and civilizations of the world knew
something about the Nature of atoms and their organization into the periodic table of the
elements, their ensuing compounds and how they form the basic structures of life, the
primordial elements that first came into existence from a big explosion of light, and how they
were made into the life elements by stars. We start with space and time itself:
Inertia is what describes mass, it is to say mass is that which resists a force; the more of it, the
more it resists a force. Let us say this mass, a sphere, is a proton. And, force A is applied to it
(Fig. 1). Then there will be a counter force B that resists change in the proton’s motion
proportional to the normal force . We will say this sphere is a hypersphere: a four dimensional
bubble whose cross-section is a three dimensional sphere, the proton. We visualize this as a
circular cross-section of a three dimensional sphere.
We suggest its resistance to change
in position is a measure of energy
over time (Planck constant, h) the
Universal constant of gravitation,
which is a measure of force over
distance, the speed of light (c the
velocity at which we move through
time), and its radius ( ) and mass
( ), and surface area ( ). We
find (Appendix 4):
Equation 38.
Where alpha is the fine structure constant:
Equation 39.
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared. Equation one can be in units of proton-seconds or just seconds because
has units of mass times time and while dividing through by the mass of a proton gives a
number of protons, the can be considered to cancel with that mass leaving pure number if
divided by one second. That number is six to about one place after the decimal.
ℕ
r
p
m
p
4π r
2
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton − secon ds
α
2
=
U
e
m
e
c
2
h 4π r
2
p
Gc
m
p
m
p
of 32 58
Six protons is carbon, the core element of biological life. We can write:
Equation 40.
Equation 41.
Where , and . Thus at one second we have carbon and at six
seconds we have hydrogen. The element hydrogen is the first element in the periodic
table has 1 proton and is the element that combines with carbon to make hydrocarbons,
the skeletons of life chemistry (As in Fig. 2).
Since hydrogen is element one it ionizes as
and since carbon is element six in group 14 and
wants four electrons to attain noble gas electron
configuration it gains four electrons and
combines with to make neutral methane:
Or, combines with two carbons covalently and
two hydrogens ionically to make a neutral
hydrocarbon chain such that it again has noble
gas electron configuration.
Interestingly, t > 6 we have fractional protons and t<1 we have we have the rest of the
elements except for a few in-between, but hydrogen and carbon are the upper and lower
limits for integer time values to produce non-fractional protons. We can see this in the
output from the program in C in Appendix 2.
1
t
1
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 6protons
1
t
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1proton
t
1
= 1second
t
2
= 2secon d s
H
+
4H
+
4H
+
+ C
4−
= CH
4
of 33 58
Thus with six-fold symmetry describing the hydrocarbons and the periodic table of the
elements being periodic over 18 groups we write:
And this, is the arithmetic breakdown one uses when playing in a meter of six
throughout the world since ancient times from Spain’s Flamenco to Middle Eastern, to
Persian to African. We write:
Eq. 42
We have predicted the radius of the proton accurately as based on carbon at t=6 seconds
in that it is experimentally:
Our data used so far is:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
But what is this unit of a second that makes this duration so descriptive of the
connection of hydrogen to carbon that makes hydrocarbons the skeletons of life
chemistry? We have more precisely:
6 ⋅ 3 = 18
9 ⋅ 2 = 18
3 + 3 + 3 = 9
2 + 2 + 2 = 6
2 ⋅ 3 = 6
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.28857 × 10
−16
m = 8.29f m
r
p
= 0.833
±
0.014 f m
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G : 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,459m /s
α : 1/137
of 34 58
Equation 43
We can actually formulate this differently than we have. We had!
!
!
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2
are lower and upper limits in an integral, then we have:!
Equation 44 !
This Equation is the generalized equation we can use for solving problems.!
Essentially we can rigorously formulate the notion of proton-seconds by considering!
Equation 45 !
Is protons-seconds squared where current density is and ( can also be
). We say!
Equation 46 !
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is!
Equation 47 !
Dividing Equation 45 through by t:!
1
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
1
t
1
1
α
2
m
p
h4π r
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h4π r
2
p
Gc
= 1proton
1
α
2
m
p
h4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z)d xd y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρdV
ℕ =
1
α
2
m
p
h4π r
2
p
Gc
of 35 58
Equation 48 !
!
Which is proton-seconds. Dividing through by t again:!
Equation 49 !
We see that if where and then J is I/m2 (current per square
meter) is analogous to amperes per per square meter which are coulombs per second through
a surface. Thus we are looking at a number of protons per second through a surface. Thus we
write:!
!
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:!
Equation 50 !
So as an example,…!
!
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter.!
!
!
#
1
α
2
m
p
h4π r
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z)d xd y
1
α
2
m
p
h4π r
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h4π r
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h4π r
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
protons
secon d
J (x, y, z) = (0,0,J ) = − J
k
d
S = d xd y
k
J ⋅ d
S = (0,0,J ) ⋅ (0,0,d x dy) = − Jd x dy
of 36 58
The Unit of a Second We need to consider the translational kinetic energy of the Earth (See
Appendix 2). 1AU=1.496E11m and . 1
year=3.1558E7s and, . m=5.972E24kg. Here we would use:
We find the same for the moon. The moon is 3.85E8m on average from the Earth center.
. Its orbital period T, is . We have
. We have the translational kinetic energies of the Earth and
moon are:
Earth:
Moon:
I found we don’t have to add on the rotational kinetic energies to the translational kinetic
energies because they are too small to have an effect within the accuracies we are working. What
we do now is create a sort of Planck’s constant h for the moon by taking its kinetic energy over
the time for a complete orbit around the earth:
And we divide this by the Earth’s kinetic energy to get:
This is the interesting thing here, we don’t need to include the rotational kinetic energies and
and one second is obtained by dividing our result by . That is:
2π (1.496E11m) = 9.3996E11m eters /year
v = 2.9785E4m eters /secon d
E =
1
2
mv
2
2π (3.85E8m) = 2.419E 9m
T = 27.32d a ys = 2.36E6s
v = 1.025E 3m /s ≈ 1000m /s
E = 2.649E 33Joules
E = 3.67E 28 Joules
h = (3.67E 28J )(2.36E6s) = 8.6612E 34J ⋅ s
8.6612E 34J ⋅ s
2.649E 33J
= 32.696secon ds
π
3
of 37 58
and
This makes sense. Because we need to bring in the lunar orbital period of 27.32 days, which
defines the month for which there are twelve in a year as compared to the Earth’s rotation which
is once per day. We have
This is close to . But the calendar month is 30 days, except for February, which is 28 days, and
29 every four years for leap year, and January, March, May, July, and August are 31 days. is
almost exactly the 31 day month that occurs 5 months a year. The leap year is every four years
because the Earth year is not exactly 365 days but is 365.25 days.
Let us now call upon equation 3 in section 1.0:
And write it
We have:
And we finally write:
Where 6 protons is carbon the basis of life. We might write it:
In order for this last equation to work perfectly the earth day has to be shorter. And indeed a
long time ago it was. But how long ago? Our Equation becomes:
32.696secon ds
π
3
= 1.05s ≈ 1secon d
π
3
= 31.006
1
π
3
= 0.03225
27.32d ays
1d a y
= 27.32
π
3
π
3
1
t
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s
1
α
2
m
p
h 4π r
2
p
Gc
= (6proton s)(1secon d )
1secon d ≈
(K . E . Moon)(Lu n arOrbital Per iod )
LunarMonth
EarthDay
(K . E . Ear th)
1
α
2
m
p
h 4π r
2
p
Gc
≈ (6proton s)
(K . E . Moon)(Lu n arOrbital Per iod )
LunarMonth
EarthDay
(K . E . Ear th)
1
α
2
m
p
h 4π r
2
p
Gc
≈ ℂ
(K . E . Moon)(Lu n arOrbital Per iod )
LunarMonth
EarthDay
(K . E . Ear th)
of 38 58
(24hours)/(1.2)=20 hours/day
Over the past several billion years the length of the year has not deviated much from 365.25 days
because of Kepler’s period for the orbit of a planet but, because the earth loses energy to the
moon, its days becomes longer over time by 0.0067 hours per million years which, we can see
from examining sedimentation band growth, which follows the lunar month. To get our 20
hours per day we have to go back about 600 million years:
24-20=0.0067t
t=597 million years
What was going on then? It turns out it was a very important time, when the Earth went through
a drastic change which lead to an explosion of life, and we started to see more complex life.
Let’s talk about what is going on here:
The reason is there. I am showing the unit of a second comes from the orbital mechanics of
the Earth-Moon-Sun System, the orbital period of the moon to the rotation period of the earth
is an approximation to pi cubed, The lunar month is determined by the orbital period of the
moon. The lunar month can be 32 days. This divided by the one day rotation of the earth is 32
is approximately pi cubed. In the equation when the lunar orbital period equals the lunar
month, they cancel and leave the kinetic energy of the moon divided by the kinetic energy of
the earth times the earth day. For it to work perfectly it means that equals one second, but it is
1.2 seconds so the earth day needs to be shorter, which it was by the right amount about 600
million years ago. That is how I got that time in earth history, which seems to be when the
earth went through a drastic change leading to complex life, according to the geologic record.
Which works well for the equation because this is a beginning time for complex life, and it
seems to be the beginning time for the synced calendar in terms of the equation. We don’t need
to put pi there because we have what we need to explain one second in terms of the Earth-
Moon-Sun Orbital mechanics. Google writes for the time 600 million years ago:
“A global ice age over 600 million years ago dramatically altered the face of the planet,
leaving a barren, flooded landscape and clear oceans, according to a study that may have
important implications for the evolution of complex life."
“The ancient sea of 600 million years ago was not soupy, the researchers said. Instead, it had
a consistency and nutrient level similar to today's oceans. This would have had major
implications for the story of animal evolution.”
We can take this further.
We needed it to be one second so the earth day had to be shorter. We went back in time to the
point where the earth day gave 1 second because 1 second is carbon giving the radius of a proton.
1
α
2
m
p
h 4π r
2
p
Gc
≈ (6proton s)
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
= 1.2secon d s
π
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6
(K . E . Moon)
(K . E . Ear th)
(Ear th Da y) = 1.2secon d s
of 39 58
I would suggest this time to be the starting point of a cosmic calendar. It turned out to be 597
million years ago an age which is perfect for such an idea because this was the Cambrian age of
the famed Cambrian explosion where the earth underwent some kind of drastic change leading
to an explosion of complex life. With carbon the core element of biological chemistry and our
equation equal to one second that gives the radius of the proton it couldn’t work better.
But, we ask what other age was pivotal to human development. It would be the end of the
Cretaceous Period when the dinosaurs went extinct 65 million years ago due to an asteroid
hitting the earth in Yucatan. I say this because we know the extinction of the dinosaurs gave
small mammals the opportunity to evolve into humans. To find out how long the earth day was
65 million years ago we write:
24-x=0.0067t
And find
x=23.5645 hours
For the length of an earth day. That is it was less by1-0.5645=0.4355 of an hour. This is close to
. We have already established that the zero of our calendar was
when when the earth day was 20 hours during the Cambrian explosion. 21 hours for an epoch
would give:
24-21=0.0067t with t=447.76 million years is approximately 450 million years
24-22=0.0067t with t=298.5 million year is approximately 300 million years
24-23=0.0067t with t=149.25 million years is approximately 150 million years.
24-23=1 and is . We have
We say 20 hours+3 hours is 0 hours+3 hours since 20 hours is the zero. We have
=dinosaur extinction
=
The 0.5645 hours beyond 23 hours is , . Where is
the radius of an equilateral triangle to its side. Thus we see a correlation between the end of the
dinosaurs and the beginning of mammalian life in our calendar zeroed with the Cambrian
Explosion. I am beginning to think there is some kind of a Natural Cosmic Calendar. If we guess
at what the pattern is here we might guess after the dinosaurs went extinct and mammals could
evolve towards humans, then the next term in our equation might be . In which case we
have:
=20+3+0.57735+0.4714=24.04875 hours
(1/2)cos30
∘
= 0.4330 = 3 /4
cosθ = 1
θ = 0
∘
cos(0
∘
) −
2
3
cos(30
∘
) =
1
2
cos(30
∘
)
3cos(0
∘
) +
2
3
cos(30
∘
)
3hours +
3
3
hours
cosθ = 0.5645 ≈ 1/ 3
θ = 54.7356
∘
1/ 3
2/3
20hours + 3hours +
3
3
hours +
2
3
hours
of 40 58
Is exactly the length of the Earth day in present times to one place after the decimal.
Can we go back in our cosmic calendar to periods before the Cambrian explosion. 65 million
years ago of course is not a precise date for the very day on which an asteroid hit earth bringing
about the extinction of dinosaurs, but given the vast expanse of time we are working with
percentage-wise that estimate is very good for our purposes. Our calendar starts at the
Cambrian explosion. What major event existed before that? It would seem about 2 billion years
ago plants and plankton came into extraordinary abundance, and these photosynthesizers
converted the primordial CO2 in the atmosphere, making it into oxygen, which they continue to
do to this day. As a result, about two billion years ago the Earth atmosphere was converted to
one having a high percentage of oxygen gas (O2). Thus a new kind of organism came into
existence, one that gets its energy from food by burning it in oxygen.
Today many organisms use copper (Cu2++) for its metabolism because it is much more soluble
than iron in its reduced state. Iron is quickly oxidized to its ferric state and in this state is
extremely insoluble. Thus, before the oxygen rich atmosphere, organisms could use iron.
The interesting thing is we don’t have to look just at the paleontological record to ascertain the
history of life life on Earth, we can find the metabolic pathways common to all life today and we
have found in common all life has certain things. We have found one of those things is an iron-
based metabolism. Thus we have a way of finding when the common ancestor to all life existed.
We call her LUCA. Since she had an Iron based metabolism she existed at least before the
explosion of photosynthesizers, which means she is at least 2 billion years old. But we have a
fossil record that goes back 3.5 billion years for the first life, and everyday that keeps getting
pushed back, even to 4 billion years, or more. The Earth and Sun formed about 5 billion years
ago.
So we may not have the data to take our calendar back to a date for LUCA. But perhaps we can
call 2 billion years accurate for the explosion of photosynthesizers and the arrival of an oxygen-
rich atmosphere.
Giordano’s Relationship And Proton-Seconds We have that a crucial value is:
From section 1.0. And we have from section 2.0 equation 13 is:
One gram of protons is:
If we are to integrate over hydrocarbon, that is from t=1 is carbon to t=6 is hydrogen, we
have:
V
c
=
gra m
proton
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
6.02E 23
m ol ⋅ protons
g
⋅
1g
m ol
= 6.02E 23pr oton s
of 41 58
Multiply that by one second from our equation in section 3.0, which is
We have
This number of 30 protons can be taken as the number 30 because
Can be taken as six seconds because, as we pointed out earlier, it is a mass divided by a
mass that can be considered to cancel with one another leaving seconds. Thus,…
1
α
2
m
p
h4π r
2
p
Gc
= 6
6
∫
6
1
dt
t
3
= − 6
(
1
6
− 1
)
= 5
∬
S
J ⋅ d
S = 5
proton s
secon d
1
α
2
m
p
h 4π r
2
p
Gc
≈ (6proton s)
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
≈ 1secon d
6
∬
S
J ⋅ d
S
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
= 30proton s
1
α
2
m
p
h4π r
2
p
Gc
= 6proton − secon ds
m
p
6
∬
S
J ⋅ d
S
(K . E . Moon)(Ear th Da y)
(K . E . Ear th)
= 30
Ear th Re volut ion s
Lu n arOrbit
of 42 58
Transform Mathematics: The transformation that rotates counter-
clockwise where is given by the standard matrix
Equation 51.
We suggest there is an aspect of Nature founded on six-fold symmetry, the example of
which we are interested in here is The Periodic Table of the Elements, because it has 18
groups which we can define by carbon, C. This because we have the following scenario:
Equations 52.
And, we pull out the 2 and the 3 and write (Fig. 3)
, ,
Where , such that
which is
given by
In general
Equations 53.
,
, ,
T : ℝ
2
→ ℝ
2
T(
x ) = A
x
A =
(
cos(θ ) −sin(θ )
sin(θ ) cosθ
)
3 + 3 + 3 = 9
2 + 2 + 2 = 6
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
2cos
π
4
= 2
2cos
π
5
=
5 + 1
2
2cos
π
6
= 3
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
f (n) = 2cos
π
n
n = 4,5,6
π
4
= 45
∘
π /5 = 36
∘
π
6
= 30
∘
Fig. 3 Dividing
line in golden
mean
of 43 58
And these can be mapped by the matrix A onto a linear vector
space (Fig. 4)
=
=
=
A =
(
2cos(θ ) −2sin(θ )
2sin(θ ) 2cosθ
)
A =
(
2cos(30
∘
) −2sin(30
∘
)
2sin(30
∘
) 2cos(30
∘
)
)
A
e
1
=
3 −1
1 3
(
1
0
)
= 3 + 1
A
e
1
=
3 −1
1 3
(
0
1
)
= 3 − 1
A =
(
2cos(36
∘
) −2sin(36
∘
)
2sin(36
∘
) 2cos(36
∘
)
)
Φ −
1
2
(5 − 5)
1
2
(5 − 5) Φ
A
e
1
= Φ +
1
2
(5 − 5)
A
e
2
= Φ +
1
2
(5 + 5)
A =
(
2cos(45
∘
) −2sin(45
∘
)
2sin(45
∘
) 2cos(45
∘
)
)
2 − 2
2 2
A
e
1
= 2 2
A
e
2
= 0
Fig. 4!
The point (1,0) can be
rotated through an
angle .
θ
of 44 58
Our is based on the square (Fig. 5)
is the line . The reflection through is given by:
Equation 54.
And our is the equilateral triangle:
To transform the square into the equilateral triangle we expand the square of base
with the matrix
To transform the square into the equilateral triangle we expand the square of base
with the matrix
And we see becomes and we have added half the square to itself. (Fig. 6)
2cos
π
4
π
4
= 45
∘
x
2
= x
1
x
2
= x
1
A =
(
0 1
1 0
)
2cos
π
6
e
1
e
1
A =
(
3/2 0
0 1
)(
1
0
)
=
3
2
e
1
3/2
Fig. 5
of 45 58
Or, better we can use the contraction
We draw in the diagonal of the the half-square and reassemble to the the two half-
triangles into an equilateral triangle. To get we take the half square and draw in the
circle of radius 1/2. (Fig. 7) We have
A =
(
1/2 0
0 1
)(
1
0
)
=
1
2
Φ
1
2
2
+ 1
2
=
1
4
+
4
4
=
5
2
5
2
+
1
2
=
5 + 1
2
= Φ
Fig. 6
Fig. 7
of 46 58
Thus we see the periodic table is 18 groups (Fig. 8).
Carbon is in group 14. We have 18-14=4 valence electrons. Hydrogen is neither a metal
or a non-metal but ionizes like a metal by losing one electron becoming and carbon
being means it needs 4 positive ions to be neutral meaning it combines with 4
hydrogens to each C, or with two hydrogens to a C and a C in long chains
(hydrocarbons) which form the Skeltons of organic compounds in life chemistry (Fig.
9) .
We have written a program that determines the protons for time (See Appendix 3) :!
24.1199 protons 0.250000 seconds 0.119904 decpart !
12.0600 protons 0.500000 seconds 0.059952 decpart !
8.0400 protons 0.750000 seconds 0.039968 decpart !
6.0300 protons 1.000000 seconds 0.029976 decpart !
4.0200 protons 1.500000 seconds 0.019984 decpart !
3.0150 protons 2.000000 seconds 0.014988 decpart !
2.1927 protons 2.750000 seconds 0.192718 decpart !
2.0100 protons 3.000000 seconds 0.009992 decpart !
1.2060 protons 5.000000 seconds 0.205995 decpart !
1.1486 protons 5.250000 seconds 0.148567 decpart !
1.0964 protons 5.500000 seconds 0.096359 decpart !
1.0487 protons 5.750000 seconds 0.048691 decpart !
1.0050 protons 6.000000 seconds 0.004996 decpart !
0.2487 protons 24.250000 seconds 0.248659 decpart !
0.2461 protons 24.500000 seconds 0.246121 decpart !
0.2436 protons 24.750000 seconds 0.243635 decpart !
H
+
C
4−
Fig. 8
Fig. 9
of 47 58
And we have angles associated with the elements from their geometry, we have a the equation
of a wave:!
Equation 55. !
That equation 6 is 1 second allows us to multiply time with special angles to get special ratios.
If we say is the amplitude and t=1 second is carbon, then trying
degrees from our triangles and squares:!
!
!
!
And for hydrogen with a value of 6 seconds, is:!
!
!
!
This is interesting to me because hydrogen being the basis of the periodic table is what it
should be, oscillating between -1 and 1 with the triangle and is 0 for the square. Carbon being
the 6 factor crux of the periodic table and core element of life chemistry is the dynamic ratios
involving the square root of three, square root of two, and one half.!
Solar System Radius
In so far as
relates carbon=1second to the Earth-Moon-Sun orbital mechanics and to the radius of a
proton through six-fold symmetry:
can we bring in our relationships
,
,
A = A
0
cosωt
A
0
= 1
ω = 30,60,45
A(60) = cos(60 ⋅ 1) = 0.5
A(30) = cos(30 ⋅ 1) = 3 /2
A(45) = cos(45 ⋅ 1) = 2 /2
A(60) = cos(60 ⋅ 6) = 1
A(30) = cos(30 ⋅ 6) = − 1
A(45) = cos(45 ⋅ 6) = 0
1
α
2
m
p
h4π r
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear thDay ≈ 1second
1
α
2
m
p
h4π r
2
p
Gc
= 6
f (n) = 2cos
π
n
n = 2,3,4,5.6.,...
f (2) = 2
f (6) = 3
of 48 58
to expand the Earth-Moon-Sun orbital mechanics to the size of the Solar System? We
consider:
Equations 56.
We move down from carbon in the periodic table to silicon (Si) and down from there to
germanium (Ge). Their densities are Si=2.33 g/cm3 and Ge=5.323 g/cm3. We have
0.21Si+0.78Ge=4.64124 g/cm3. Consider this the starting point for the density of a thin
disc decreasing linearly from the Sun to Pluto (49.5 AU=7.4E14cm).
Equation 57.
=
The sum of the masses of the planets is 2.668E30 grams. The accuracy is:
We have the radius of the solar system is given by:
Equation 58.
Interestingly, the relative abundances of Nitrogen and oxygen are 0.78 and 0.21 in the
Earth air. In fact, the molar mass of air as a mixture is:
Interestingly as well, by molar mass
6
∫
3
2
1
t
2
dt = 6
(
1
2
−
1
3
)
= 0.78
∫
2
3
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
=
π − 4
2 2 = 0.21
= 0.21
M =
∫
2π
0
∫
R
0
ρ
0
(
1 −
r
R
)
rdrdθ =
πρ
0
R
2
3
π(4.64124)(7.4E14)
2
3
= 2.661E30g
2.66
2.668
100 = 99.736
R
s
=
3M
p
π(0.78Ge + 0.21Si )
air = 0.78N
2
+ 0.21O
2
= 29.0g/m o l
of 49 58
Carbon equals one second is the radius of a proton:
The experimental radius of a proton is:
In our integral:
where did come from? We had Equations
,
We write and get and
Abundances of Primordial Elements If we can explain the Earth-Moon-Sun orbital
mechanics, the radius of the solar system, and the radius of the proton, can we predict
the relative abundances of the primordial elements created at the beginning of the
Universe - hydrogen and helium - from which all of the elements were made by stars in
nucleosynthesis?
We consider a gaussian wave packet at t=0:
We say that d which in quantum mechanics would be the delocalization length when
squared is
air
H
2
O
≈ Φ
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 8.29f m
r
p
= 0.833
±
0.014 f m
∫
2
3
cos
−1
(x /2)d x =
1
6
(
3π − 6
)
= 0.21
cos
−1
(x /2)
f (n) = 2cos
π
n
n = 4,5,6
x = 2cosθ
θ = cos
−1
(x /2)
cos
−1
( 2/2) = 45
∘
=
π
4
cos
−1
( 3/2) = 30
∘
=
π
6
ψ (x,0) = Ae
−
x
2
2d
2
of 50 58
!
We write the wave packet as a Fourier transform:!
!
We use the identity that gives the integral of a quadratic:!
!
!
Calculate the Gaussian integral of dp:!
and. !
Equation 59.!
!
!
(
Si − C
C
)
2
=
16
9
ψ (x,0) = Ae
−
x
2
2d
2
=
∫
dp
2π ℏ
ϕ
p
e
i
h
px
∫
∞
−∞
e
−α
2
x+βx
d x =
π
α
e
β
2
4α
ψ (x, t) =
∫
dp ⋅ e
p
2
d
2
2ℏ
2
⋅ e
−
i
h
( px−
p
2
2m
t)
α =
d
2
2ℏ
2
+
it
2mℏ
β =
i x
ℏ
ψ
2
= exp
[
−
x
2
d
2
⋅
1
1 + t
2
/τ
2
]
τ =
m d
2
ℏ
ψ
2
= exp −
C
2
x
2
(Si − C )
2
⋅
1
1 +
[
ℏC
2
m(S i − C)
2
]
2
t
2
ψ
2
= exp −
9
16
x
2
⋅
1
1 +
ℏ
2
81
m
2
256
t
2
of 51 58
For Hydrogen (t=6 seconds):
=74%
For Helium (t=3 seconds):
=26%
This is in close agreement with what we observe. The Universe is 74% hydrogen and
24% helium, The remaining 2% is comprised by the rest of the elements.
ψ
2
= (1)exp −
9
16
(1proton)
2
⋅
1
1 +
(0.075)81
(1)256
(6seconds)
2
ψ
2
= (2)exp −
9
16
(2proton)
2
⋅
1
1 +
(0.075)81
(4)256
(3seconds)
2
of 52 58
Appendix 2
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
of 53 58
Appendix 3
Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by
the mass of a proton. But these masses can be considered to cancel and leave pure number. We
make a program that looks for close to whole number solutions so we can create a table of values
for problem solving.
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?):
");
scanf("%i", &n);
}
while (n>=101);
1
α
2
m
p
h 4π r
2
p
Gc
of 54 58
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
of 55 58
Appendix 4
We need offer an interpretation of time, as well. (Illustration next page)
In The Time Machine, a science fiction story by HG Wells, the time traveller describes
time as physical distance, the direction through which the universe is falling at the speed
of light, c. Thus, not only when we move through space do we travel through a distance
at a velocity v, but we travel through a distance t at a velocity c. If we draw the picture
and account for that distance and velocity as well, we arrive at time dilation as given by
relativity theory. He writes:
I think that at the time none of us quite believed in the Time Machine. The fact is, the
Time Traveller was one of those men who are too clever to be believed: you never felt
that you saw all around him; you always suspected some subtle reserve, some
ingenuity in ambush, behind his lucid frankness.
Since we don’t experience our motion through time as we fall through it with the
universe, we change the sign in . And have the relativistic equation for time
dilation:
x
2
+ t
2
= d
2
x = vt
2
0
v
2
t
2
0
c
2
t
2
+
c
2
t
2
0
c
2
t
2
= 1
t
2
0
(
v
2
c
2
+ 1
)
= t
2
t
0
=
t
1 +
v
2
c
2
v
2
/c
2
t
0
=
t
1 −
v
2
c
2
of 56 58
#
of 57 58
5.6 Bringing It All Together
We take equation 8
And Equation 37
And write
And equation 35
And equation 43
And write
μ =
π
8α
2
⋅
m
e
e
6
c
2
ϵ
3
0
h
4
= 2.39E − 29kg /m
k =
1
m
2
i
⋅ h
(1 + α)
G
⋅ N
A
𝔼
μ
k
= 1.912E − 26kg/s
k v
e
= 6
1
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.004996352secon d s
1
α
2
m
p
h4π r
2
p
Gck v
e
= 1proton ⋅ secon d
of 58 58
The Author
